Probability of exactly one defective unit Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random. What is the probability that exactly one unit is defective? * * * My answer would be $P(\text{Defect}=1) = P(\text{Defect})\times P(\text{Not defect})\times P(\text{Not defect}) = 5/100 \times 95/99 \times 94/98$ However, I am not sure whether or not this is correct or not. Can someone verify?

Your answer should be $$\frac{\binom{5}{1}\binom{95}{2}}{\binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.